Horizon and swath
From a satellite at radius r = R + h, a ray that just grazes the sphere meets the surface at a right angle. The nadir angle to that geometric limb and the Earth-central angle are
ηhorizon = arcsin(R / (R+h)) λhorizon = arccos(R / (R+h))
The horizon ground range is R λ, and the horizon-to-horizon swath is twice that. At 400 km over a 6,371 km mean sphere, ηhorizon is about 70.21°, λhorizon about 19.79°, and the horizon radius about 2,200 km. For a look (or half-FOV) η inside the limb, the standard spherical identity is
λ = arcsin( ((R+h)/R) sin η ) − η
The ground range from the subsatellite point is R λ, and a symmetric nadir field of view has swath width 2 R λ. A 15° half-angle at 400 km covers about 0.97° of Earth-central angle, or roughly 107 km of ground range (a 215 km swath). A look past the limb has no intersection.
Period and subsatellite speed
For context, the circular-orbit period and inertial speed use the same Earth gravitational parameter as the orbital velocity and period calculator: T = 2π √(r³/μ) and V = √(μ/r), with μ = 3.986004418×10¹⁴ m³/s². A 400 km circular orbit is about 92.4 minutes. The speed of the subsatellite point on a non-rotating Earth is V × R/(R+h). A true ground track on the rotating Earth also depends on inclination and nodal motion, which are not modelled here.
Sources
Vallado, Fundamentals of Astrodynamics and Applications (Earth central angle, horizon, swath); Wertz and Larson, Space Mission Analysis and Design; NASA / JPL planetary constants (μ and mean radius). For circular, escape, and vis-viva speeds, see orbital velocity and period.
Frequently asked questions
How far is the horizon from 400 km altitude?
About 2,200 km of ground range on a 6,371 km mean sphere (λ ≈ 19.79°, η ≈ 70.21°). The horizon-to-horizon swath is twice that, about 4,400 km.
What swath does a 15° half-angle give at 400 km?
About 215 km (a ground range of about 107 km either side of nadir). The Earth-central angle is λ = arcsin(((R+h)/R) sin η) − η ≈ 0.97°.
Is the reported ground speed a GPS ground-track speed?
No. It is the inertial circular speed scaled by R/(R+h), as if the Earth were not rotating. A ground track on the rotating Earth also depends on inclination and the satellite’s heading relative to the equator.